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Dirac equation in natural units is: $$\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi=0$$ where $\gamma^{0}=\pmatrix{I_{2} & 0\\ 0 & -I_{2}}$ and $\gamma^{n}=\pmatrix{0 & \sigma_{n}\\ -\sigma_{n} & 0}$, with $n=1,2,3.$

Gamma matrices must have the trace zero and the number of positive components equal to the number of negative components. Using Pauli matrices, the Dirac equation can describe spin 1/2 particles. I am wondering if there is a deeper reason for using Pauli matrices besides because the math works and they satisfy those conditions above.

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  • $\begingroup$ Dirac didn't come up with this idea. He derived a particular representation of a particular Clifford algebra which had i.e. MUST contain the Pauli matrices in them. $\endgroup$ – DanielC Oct 31 '17 at 16:25
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Dirac wanted to reformulate a relativistic version of the Schrödinger equation without the problems arising in the Klein-Gordon equation (for example, the fact that the probability density is negative in the KG equation). So he modified the KG equation $$(\partial_\mu\partial^{\mu} + m^2)\psi=0$$ to the first order equation $$(i\gamma^\mu\partial_{\mu}-m)\psi=0$$ which can be proved that gives a positive definite probability density. But what are these coefficients $\gamma^\mu$? Applying $i\gamma^{\nu}\partial_{\nu}$ on both sides of the equation gives $$-\gamma^{\nu}\gamma^{\mu}\partial_{\nu}\partial_{\mu}\psi=im\gamma^{\nu}\partial_{\nu}\psi=m^2\psi$$ Where in the last step we have used the result of the original Dirac equation. Now, we can write the last equation as: $$-\frac{1}{2}(\gamma^{\nu}\gamma^{\mu}+\gamma^{\mu}\gamma^{\nu})\partial_{\nu}\partial_{\mu}\psi=m^2\psi$$ We see that the term in the parenthesis is the well known anticommutator $\{ \gamma^{\nu},\gamma^{\mu} \}=\gamma^{\nu}\gamma^{\mu}+\gamma^{\mu}\gamma^{\nu}$. We know as well that if we have a plain wave $\psi\sim e^{ip^{\mu}x_{\mu}}$ and since $p_{\mu}p^{\mu}=m^2$ the Dirac equation will become: $$\frac{1}{2}\{ \gamma^{\nu},\gamma^{\mu} \}p_{\nu}p_{\mu}=m^2$$ It's easy to see that in order to fulfill the previous equations, the anticommutator has to be: $$\{ \gamma^{\nu},\gamma^{\mu} \}=2\eta^{\nu\mu}$$ Since $\eta^{\nu\mu}$ are the components of a $4\times4$ matrix, we can easily deduce that the four quantities $\gamma^\mu$ have to be matrices as well. The above relation already defines the $\gamma^{\mu}$, but we will see some other nice properties. The easy ones are: $$(\gamma^0)^2=\eta^{00}=1, \ \ (\gamma^i)^2=-1, \ \ \gamma^0\gamma^i+\gamma^i\gamma^0=0, \ \ \gamma^i\gamma^j+\gamma^j\gamma^i=2\delta^{ij}$$ From the first one we can deduce that the eigenvalues of $\gamma^0$ are $+1$ and $-1$. We have also that if $\nu\neq\mu$: $$\gamma^{\mu}\gamma^{\nu}=-\gamma^{\nu}\gamma^{\mu}$$ In particular, if we take $\nu=0$ and $\mu=i$ we get: $$\gamma^0\gamma^i=-\gamma^i\gamma^0, \ \text{multiplying by $\gamma^i$ on the right}\to\gamma^0=\gamma^i\gamma^0\gamma^i$$ If we take the trace of the last result: $$\text{Tr}(\gamma^0)=\text{Tr}(\gamma^i\gamma^0\gamma^i)=\text{Tr}(\gamma^i\gamma^i\gamma^0)=\text{Tr}(-\gamma^0)=-\text{Tr}(\gamma^0)\Longrightarrow \text{Tr}(\gamma^0)=0$$ Here we have used some properties of the trace, namely $\text{Tr}(AB)=\text{Tr}(BA)$ and $\text{Tr}(-A)=-\text{Tr}(A)$. From this fact and that the eigenvalues of $\gamma^0$ are $+1$ and $-1$ we can see that this matrix has to be of even dimension. It can't be of dimension 2 because it would not fulfill the above properties. It's not hard to see that its dimension has to be 4. Now we can look upon the Dirac equation: $$i\gamma^{\mu}\partial_{\mu}\psi=m\psi\Longrightarrow i\gamma^{0}\frac{\partial\psi}{\partial t}=i\gamma^i\frac{\partial\psi}{\partial x^i}+m\psi$$ This equation only makes sense if $\psi$ is a 4-dimensional object, because we have $4\times 4$ matrices acting on it. We call $\psi$ a spinor (4-spinor, Dirac spinor), and we will see that is defined by its transformation properties under Lorentz transformations. Recalling the Dirac equation and multiplying by $\gamma^0$ on both sides: $$i\frac{\partial\psi}{\partial t}=i\gamma^0\gamma^i\frac{\partial\psi}{\partial x^i}+m\gamma^0\psi$$ We can identify the left hand side of the equation as the hamiltonian operator $H\psi=i\frac{\partial\psi}{\partial t}$, and obviously has the form: $$H=\gamma^0\gamma^ip_i+m\gamma^0$$ Sometimes it's convenient to express it as: $$ H=\boldsymbol{\alpha}\cdot\mathbf{p}+\beta m$$ Where we define the operators $\boldsymbol{\alpha}$ and $\beta$: $$\boldsymbol{\alpha}=\left(\begin{array}{cc} 0 & \boldsymbol{\sigma} \\ \boldsymbol{\sigma} & 0 \end{array}\right), \ \ \beta=\gamma^{0}$$ By requiring to be hermitian $H=H^{\dagger}$ we have: $$(\gamma^0\gamma^ip_i)^{\dagger}+(m\gamma^0)^{\dagger}=\gamma^0\gamma^ip_i+m\gamma^0$$ Since the momentum is already hermitean $p_i=p_i^{\dagger}$ we have the conditions: $$(\gamma^0)^{\dagger}=\gamma^0, \ \ (\gamma^0\gamma^i)^{\dagger}=\gamma^0\gamma^i$$ From the first one and since $(\gamma^0\gamma^i)^{\dagger}=(\gamma^i)^{\dagger}(\gamma^0)^{\dagger}$ we see that $(\gamma^i)^{\dagger}=-\gamma^i$. One can write all this properties in a single formula: $$\gamma^0(\gamma^\mu)^{\dagger}\gamma^0=\gamma^{\mu}$$ Indeed there are not only one set of matrices that fulfill the above properties. A common way of writing them is what is known as the Dirac representation: $$ \gamma^0=\left(\begin{array}{cc} \mathbb{I} & 0 \\ 0 & - \mathbb{I} \end{array}\right), \ \ \gamma^i=\left(\begin{array}{cc} 0 & \sigma^i \\ -\sigma^i & 0 \end{array}\right) $$ Where of course $\sigma^i$ are the Pauli matrices: $$\sigma^1=\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right), \ \sigma^2=\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right), \ \sigma^3=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$$ Therefore, the Pauli matrices appear as a result of several conditions the $\gamma$ matrices must fulfill. Some other well known representations of the $\gamma$ matrices that fulfill the properties are the Weyl representation or the Majorana representation, both equally useful in their respective aspects. In fact, all the representations are equivalent up to a unitary transformation: $$\gamma^{\mu}_{_{R'}}=\mathcal{U}\gamma^{\mu}_{_R}\mathcal{ \ U}^{-1}$$ Where $\mathcal{U}$ is a unitary operator and $R$ and $R'$ label the representation.

Sorry for the long explanation, but I hope it helps!

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Within the Poincare group representation theory, the $\sigma_{\mu} = (1,\sigma)$ matrices appear as lorentz-covariant objects when we build the "up to sign" homorphism between the proper ortochronous Lorentz group $SO^{\uparrow}_{+}(3,1)$ and the $SL(2,C)$ group; namely, with $N_{ab}(\Lambda)$ being the $SL(2,C)$ transformation corresponding to the Lorentz group transformation $\Lambda_{\mu}^{\ \nu}$ and $N_{\dot{a}\dot{b}}(\Lambda)$ being the corresponding conjugated transformation, one have $$ N_{a'a}(\Lambda)\sigma_{\mu}^{a\dot{b}}N_{\dot{b}\dot{b}'}(\Lambda) = \Lambda_{\mu}^{\ \nu}(\sigma_{\nu})_{a'\dot{b}'} $$ For example, each 4-vector $A_{\mu}$ is isomorphic to the matrix $A \equiv A_{\mu}\sigma^{\mu}$.

These matrices always occur when we're dealing with the spinorial representations of the Poincare group. The latter carries half-integer spin, and particular example of such representations is Dirac representation. This is the modern picture.

However, Dirac in late twenties simply used the method "it works", since he constructed his equation from the requirement to obtain the first-order derivative equation describing the wave-function with positive probability.

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  • $\begingroup$ The tricky part is that what you wrote has nothing to do with the representation theory of the Poincare group we need in physics. You have failed to explain why the gamma/Dirac matrices contain a block-diagonal form of sigma/Pauli matrices. This is what the OP asked for. $\endgroup$ – DanielC Oct 31 '17 at 16:39
  • $\begingroup$ @DanielC : have you read the question, or (as for me) not? The question is "How did Dirac come up with the idea of using Pauli matrices?" See the example of a question corresponding to the one you've "invented": physics.stackexchange.com/questions/261893/… $\endgroup$ – Name YYY Oct 31 '17 at 19:25
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See this article for a very interesting and comprehensive discussion of this matter. http://www.mathpages.com/home/kmath654/kmath654.htm

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